Question: Simplify and expand the following expression: $ \dfrac{k + 9}{2k - 2}-\dfrac{k}{k - 4} $
Solution: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(2k - 2)(k - 4)$ Multiply the first term by $\dfrac{k - 4}{k - 4}$ $ \begin{align*} \dfrac{k + 9}{2k - 2} \times \dfrac{k - 4}{k - 4} & = \dfrac{(k + 9)(k - 4)}{(2k - 2)(k - 4)} \\ & = \dfrac{k^2 + 5k - 36}{(2k - 2)(k - 4)}\end{align*} $ Multiply the second term by $\dfrac{2k - 2}{2k - 2}$ $ \begin{align*} \dfrac{k}{k - 4} \times \dfrac{2k - 2}{2k - 2} & = \dfrac{(k)(2k - 2)}{(k - 4)(2k - 2)} \\ & = \dfrac{2k^2 - 2k}{(k - 4)(2k - 2)}\end{align*} $ Now we have: $ = \dfrac{k^2 + 5k - 36}{(2k - 2)(k - 4)} - \dfrac{2k^2 - 2k}{(k - 4)(2k - 2)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{k^2 + 5k - 36 - (2k^2 - 2k)}{(2k - 2)(k - 4)} $ $ = \dfrac{k^2 + 5k - 36 - 2k^2 + 2k}{(2k - 2)(k - 4)} $ $ = \dfrac{-k^2 + 7k - 36}{(2k - 2)(k - 4)}$ Expand the denominator: $ = \dfrac{-k^2 + 7k - 36}{2k^2 - 10k + 8}$